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{SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 50 "High School Modul
es > Geometry by Gregory A. Moore" }}{PARA 3 "" 0 "" {TEXT -1 4 "    \+
" }{TEXT 256 24 "Impossible Constructions" }}{PARA 0 "" 0 "" {TEXT -1 
74 "\nAn exploration of the three famous impossible constructions of g
eometry.\n" }}{PARA 0 "" 0 "" {TEXT 258 153 "[Directions : Execute the
 Code Resource section first. Although there will be no output immedia
tely, these definitions are used later in this worksheet.]" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 
260 7 "0. Code" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart; " 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "with(plots):   " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 294 "scan := 'axes = none, scali
ng = constrained':\n\nth2 :=  'thickness = 2':\nth3 :=  'thickness = 3
':\nl2 :=  'linestyle = 2':\nl3 :=  'linestyle = 3':\n\ncb := 'color =
 COLOR(RGB,.6,.6,.8)':\ncr := 'color = COLOR(RGB,.95,.1,.1)':\ncg := '
color = COLOR(RGB,.5,.9,.5)':\nco := 'color = COLOR(RGB,.9,.5,.0)':" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 
"" {TEXT -1 1 " " }{TEXT 261 25 "1. Possible Constructions" }}{PARA 0 
"" 0 "" {TEXT -1 162 "\nThere are many constructions possible with onl
y straight, compass, and pencil - too numerous to catalog. Here is one
 simple example of a construction.\n\n         " }{TEXT 265 19 "Bisect
ing a Segment" }{TEXT -1 208 "\n\nDraw arcs from each endpoint, where \+
the radius is greater than half of the length of the segment. Find the
 points of intersection of the arcs, and draw a line connecting them. \+
This line bisects the segment." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 71 "AB := [[-3,0],[3,0]]:\nL := plot( AB, th3, co, scan):\ndisplay( \+
L, scan);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 111 "a1 := plottoo
ls[arc]([-3,0], 4, -1.2..1.2):\na2 := plottools[arc]([3,0], 4, 2.2..4)
:\ndisplay( [a1,a2, L], scan);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 89 "AB := [[0,-3],[0,3]]:\nL2 := plot( AB, color = red, scan):\ndi
splay( [a1,a2, L, L2], scan);" }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n " }}}
{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 266 27 "2. Impossible Co
nstructions" }}{PARA 0 "" 0 "" {TEXT -1 115 "\nFrom antiquity there we
re three constructions which were long sought after but never found :
\n\n\011                   " }{TEXT 267 108 "1. Trisecting an Angle\n
\n\011                     2. Squaring a Circle\n\n\011               \+
      3. Doubling a Cube" }{TEXT -1 1215 "\n\n\nIn mathematics the wor
d impossible means something different than it means in other areas of
 study. For example, someone might say that it is impossible to turn l
ead into gold, and indeed there is no known way to do this. However, b
efore the 1940's it was equally impossible to turn hydrogen into heliu
m, and yet that is what a hydrogen bomb can now do. In the 1950's it w
as impossible to send a man to the moon. in the 1800's it was impossib
le for a man to fly, or to cross the Atlantic in less than a day. The \+
word \"impossible\" in these contexts means \"not doable at this time
\". \n\nHowever, in mathematics, the word means something different. I
t means \"not doable at this time, or any time in the future\" - a muc
h stronger statement. In the late 1800s, it was proven that all three \+
of the constructions mentioned above are impossible. This was done by \+
abstracting the processes of these constructions and using abstract al
gebra to prove the impossibility of these tasks ever being accomplishe
d.\n\nMany individuals don't understand this concept and continue to t
ry to find constructions. One author has compiled a book of many of th
ese attempts - which are, of course, doomed to failure, but make amusi
ng reading." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 242 "Because these constructions are impossible, we will not be abl
e to solve them using the straightedge and compass. However, using Map
le we can approximate to a high degree of accuracy the solutions to be
tter understand the unsuccessful quests." }}{PARA 0 "" 0 "" {TEXT -1 
2 "\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 262 32 "3. Im
possible Construction 1 :  " }{TEXT 268 16 "Trisect an Angle" }}{PARA 
0 "" 0 "" {TEXT -1 410 "\nTo trisect an arbitrary angle means to divid
e it evenly into three sections. This is impossible using the rules of
 construction. It's not impossible to get an approximation (no matter \+
how accurate). Using Maple we can use numerical methods to find the tr
isection. So we will see below the goal of trisection, although we are
 not using only straightedge and compass - because it's impossible to \+
do it that way!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 96 "OA := [[0,0],[5,2]]:\nOB := [[0,0],[3,8]]:\nL \+
:= plot( [OA,OB], th3, co, scan):\ndisplay( L, scan);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 112 "t1 := evalf(arctan( OA[2,2]/OA[2,1
]),15):\nt2 := evalf(arctan( OB[2,2]/OB[2,1]),15):\ntheta := t2-t1; `(
radians)`;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "arcbetw := pl
ottools[arc]([0,0], 3, t1..t2):\ndisplay( L,arcbetw, scan); " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 334 "u1 := t1 + theta/3;\nu2 := \+
u1 + theta/3; \n\nla := sqrt( (OA[1,1]-OA[2,1])^2  + (OA[1,2]-OA[2,2])
^2);\nlb := sqrt( (OB[1,1]-OB[2,1])^2  + (OB[1,2]-OB[2,2])^2);\nlc := \+
evalf( (2*la + lb)/3 );\nld := evalf( (la + 2*lb)/3 );\nOC := [[0,0],[
lc*cos(u1),lc*sin(u1)]];\nOD := [[0,0],[ld*cos(u2),ld*sin(u2)]];\nL2 :
= plot( [OC,OD], th2, l3, cb, scan):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 271 "t3 := evalf(arctan( OC[2,2]/OC[2,1]),15);\nt4 := eva
lf(arctan( OD[2,2]/OD[2,1]),15);\narc1 := plottools[arc]([0,0], 3.2, t
1..t3):\narc2 := plottools[arc]([0,0], 3.4, t3..t4):\narc3 := plottool
s[arc]([0,0], 3.2, t4..t2):\narcs := display( \{arc1,arc2,arc3\}, thic
kness = 5, cg):\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "displa
y( \{L,L2,arcbetw,arcs \}, scan);" }}}{PARA 0 "" 0 "" {TEXT -1 51 "\n
\nHere is another example with a different angle. \n" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 44 "OA := [[0,0],[13,4]]:\nOB := [[0,0],[1,11
]]:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 739 "L := plot( [OA,OB
], th3, co, scan):\nt1 := evalf(arctan( OA[2,2]/OA[2,1]),15):\nt2 := e
valf(arctan( OB[2,2]/OB[2,1]),15):\ntheta := t2-t1:\narcbetw := plotto
ols[arc]([0,0], 3, t1..t2):\nu1 := t1 + theta/3:\nu2 := u1 + theta/3:
\nla := sqrt( (OA[1,1]-OA[2,1])^2  + (OA[1,2]-OA[2,2])^2):\nlb := sqrt
( (OB[1,1]-OB[2,1])^2  + (OB[1,2]-OB[2,2])^2):\nlc := evalf( (2*la + l
b)/3 ):\nld := evalf( (la + 2*lb)/3 ):\nOC := [[0,0],[lc*cos(u1),lc*si
n(u1)]]:\nOD := [[0,0],[ld*cos(u2),ld*sin(u2)]]:\nL2 := plot( [OC,OD],
 th2, l3, cb, scan):\narc1 := plottools[arc]([0,0], 3.3, t1..u1):\narc
2 := plottools[arc]([0,0], 3.5, u1..u2):\narc3 := plottools[arc]([0,0]
, 3.3, u2..t2):\narcs := display( \{arc1,arc2,arc3\}, thickness = 5, c
g):\ndisplay( \{L,L2,arcbetw,arcs \}, scan);\n" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }
{TEXT 263 32 "4. Impossible Construction 2 :  " }{TEXT 269 15 "Square \+
a Circle" }}{PARA 0 "" 0 "" {TEXT -1 114 "\nTo square a circle means t
o find a square with area equal to a given circle. Lets start with a s
quare of side 10." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 150 "s := 5
:\nsq := [[-s,-s],[s,-s],[s,s],[-s,s],[-s,-s]]:\nSQ := plot( sq, th3, \+
cr, scan):\nSQ2 := plot( sq, th3, cr, scan, filled = true ):\ndisplay(
 SQ2 );\n" }}}{PARA 0 "" 0 "" {TEXT -1 37 "\nHere is the area of the s
quare, 100." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "Area := (2*s)
^2;" }}}{PARA 0 "" 0 "" {TEXT -1 126 "\n\nNow, we want to find a circl
e with the same area. We can solve the area formula for a circle, with
 area equal to 100, for r." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
76 "r :='r':   theta := 'theta';\nArea = Pi*r^2;\nsolve(%, r);\nr := s
qrt(Area/Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 123 "\nSo this is the radiu
s of the desired circle. Now we only need to graph it.  Both of these \+
figures have the same area, 100." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 87 "CIR := polarplot( r, theta = 0..2*Pi, scan, th3, cb, filled = \+
true):\ndisplay( SQ, CIR);" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " \+
" }{TEXT 264 32 "5. Impossible Construction 3 :  " }{TEXT 270 13 "Doub
le a Cube" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 197 "To double a cube means to find a cube with exactl
y twice the volumne. This is equivalent to constructing the cube root \+
of two. This is particularly tricky, since this is a three dimensional
 figure." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
272 "(We invoke a new package of commands. Ignore the long list of war
nings for new definitions. We construct a cube, compute the side we wi
ll need to double it, and then draw both together. Although it may not
 look like it, the larger cube has twice the volume often smaller!)" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 143 "with(geom3d):\npoint(orig,0,0,0):\nside_len := 8;\ncube( cube_1
, orig ,side_len/2*sqrt(3) ):\ndraw( cube_1 , axes = framed, lightmode
l = light1 );\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "`Volume \+
of the first cube of side 8` =evalf(  volume( cube_1 ));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 105 "`The side of the larger cube with \+
twice the volume is :`;\nside_2 := evalf( ( 2*volume( cube_1 ))^(1/3) \+
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "cube( cube_2, orig, s
ide_2 ):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 135 "draw(  [ cube_
1, \n         cube_2(style=wireframe, color = blue)], \n         axes \+
= framed, orientation =[30,82], shading=ZGRAYSCALE );" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n \+
" }}}{PARA 0 "" 0 "" {TEXT 259 36 "\n         \251 2002 Waterloo Maple
 Inc " }}}{MARK "0 1" 40 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }